# Equilibrium temperature

An insulated bucket contains 150 g of ice at 0°C. If 26.0g of steam is injected at 100°C into the bucket, what is the final equilibrium temperature of the system in °C?

this i what ive done so far
q(ice)=150g*333.5J/g=50025J
q(steam)=26g*2257J/g=58682J

q(ice) + q(ice-water) = -[q(steam) + q(steam-water)]

50025J + 4.18J/g*150g(Tf-273)=-[58682J + 4.18J/g*26g(Tf-373)]
20025J + 327J*Tf - 171171J = -58682J -108J*Tf + 40538J
-121146J + 627J*Tf = -18145 - 108J*Tf
735J*Tf=103002
Tf=140degree K

this is obviously wrong....what am i missing?

Doc Al
Mentor
You are adding an extra minus sign. Think of it this way:
Heat gained by ice/water = Heat lost by steam/water

Doc Al said:
You are adding an extra minus sign. Think of it this way:
Heat gained by ice/water = Heat lost by steam/water

so you are saying
q(ice) + q(ice-water) = q(steam) + q(steam-water)

50025J + 4.18J/g*150g(Tf-273)=58682J + 4.18J/g*26g(Tf-373)
121146J + 627J*Tf = 58682J + 108J*Tf - 40538J
519J*Tf=139290J
Tf=268degree K

which is still wrong......
did i understand you correctly?

Doc Al
Mentor
jennypear said:
50025J + 4.18J/g*150g(Tf-273)=58682J + 4.18J/g*26g(Tf-373)
Note that since Tf will be between 273 and 373 K, Tf-373 is negative.

I always think in terms of heat lost versus heat gained; these quantities are always positive.

Doc Al
Mentor
jennypear said:
q(ice) + q(ice-water) = -[q(steam) + q(steam-water)]

50025J + 4.18J/g*150g(Tf-273)=-[58682J + 4.18J/g*26g(Tf-373)]
If you prefer to use this form of the heat flow equation, which is perfectly OK, then be sure to use the correct signs for all quantities. Note that q(steam) should be negative since heat is flowing out of the steam.

hmm tricky:) thanks so much will look at it that way in the future